# Element has Idempotent Power in Finite Semigroup

## Theorem

Let $\left({S, \circ}\right)$ be a finite semigroup.

For every element in $\left({S, \circ}\right)$, there is a power of that element which is idempotent.

That is:

$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$

## Proof

From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$

Let $m > n$.

Let $n = k, m = k + l$.

Then:

$\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$

Now we show that:

$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$

That is, that $x^{k l}$ is idempotent.

First:

 $$\displaystyle x^k$$ $$=$$ $$\displaystyle x^{k + l}$$ $$\displaystyle \implies$$ $$\displaystyle x^k \circ x^l$$ $$=$$ $$\displaystyle x^{k + l} \circ x^l$$ Both sides ${} \circ x^l$ $$\displaystyle \implies$$ $$\displaystyle x^{k + l}$$ $$=$$ $$\displaystyle x^{k + 2 l}$$ Index Laws for Semigroup‎ $$\displaystyle \implies$$ $$\displaystyle x^k$$ $$=$$ $$\displaystyle x^{k + 2 l}$$ as $x^k = x^{k + l}$

From here we can easily prove by induction that:

$\forall n \in \N: x^k = x^{k + n l}$

In particular:

$x^k = x^{k + k l} = x^{k \left({l + 1}\right)}$

There are two cases to consider:

$(1): \quad$ If $l = 1$, then $x^k = x^{k \left({l + 1}\right)} = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
$(2): \quad$ If $l > 1$, then:
 $$\displaystyle x^k$$ $$=$$ $$\displaystyle x^{k + kl}$$ $$\displaystyle$$ $$=$$ $$\displaystyle x^{k \left({l + 1}\right)}$$ $$\displaystyle \implies$$ $$\displaystyle x^k \circ x^{k \left({l - 1}\right)}$$ $$=$$ $$\displaystyle x^{k \left({l + 1}\right)} \circ x^{k \left({l - 1}\right)}$$ Both sides ${} \circ x^{k \left({l - 1}\right)}$ $$\displaystyle \implies$$ $$\displaystyle x^{k + k l - k}$$ $$=$$ $$\displaystyle x^{kl + k + k l - k}$$ Index Laws for Semigroup‎ $$\displaystyle \implies$$ $$\displaystyle x^{k l}$$ $$=$$ $$\displaystyle x^{k l + k l}$$ $$\displaystyle$$ $$=$$ $$\displaystyle x^{k l} \circ x^{k l}$$

... and again, $x^{k l}$ is idempotent.

$\blacksquare$